Is this equation true for positive values of x and y?

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SUMMARY

The equation $y^2+x*\sqrt{4x^2+y^2}=y^2+\frac{x}{y}*\sqrt{4(\frac{x}{y})^2+1}$ is confirmed to be true for positive values of x and y. The transformation of the left-hand side to $y^2+x \cdot |y| \cdot \sqrt{ \frac{4x^2}{y^2}+1}$ demonstrates that both sides are equivalent when y is greater than zero. This confirms the validity of the equation under the specified conditions.

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cbarker1
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Hello,

I need some aid to see if this true:

$y^2+x*\sqrt{4x^2+y^2}=y^2+\frac{x}{y}*\sqrt{4(\frac{x}{y})^2+1}$ provide that y>0 and x>0.Thank you,

Cbarker1
 
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Cbarker1 said:
Hello,

I need some aid to see if this true:

$y^2+x*\sqrt{4x^2+y^2}=y^2+\frac{x}{y}*\sqrt{4(\frac{x}{y})^2+1}$ provide that y>0 and x>0.Thank you,

Cbarker1

(Wave)

No. It holds that $y^2+x \cdot \sqrt{4x^2+y^2}=y^2+ x \cdot \sqrt{y^2 \left( \frac{4x^2}{y^2}+1\right)}=y^2+x \cdot |y| \cdot \sqrt{ \frac{4x^2}{y^2}+1}$.And provided that $y>0$, $y^2+x \cdot |y| \cdot \sqrt{ \frac{4x^2}{y^2}+1}=y^2+x \cdot y \cdot \sqrt{ \frac{4x^2}{y^2}+1}$
 
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